A history class is comprised of 5 female and 8 male students. If the instructor of the class randomly chooses 8 students from the class for an oral exam, what is the probability that 2 female students and 6 male students will be selected? Round your answer to decimal places.

To find the probability, we need to determine the total number of ways to choose 8 students from a class of 13, and the number of ways to choose 2 female students and 6 male students.

To find the total number of ways to choose 8 students from 13, we can use the combination formula, which is denoted as:

C(n, r) = n! / (r!(n-r)!)

Where n is the total number of students in the class (13) and r is the number of students to be chosen (8).

Let's calculate:

C(13, 8) = 13! / (8!(13-8)!) = 13! / (8!5!) = (13 * 12 * 11 * 10 * 9) / (5 * 4 * 3 * 2 * 1) = 1287

So, there are 1287 ways to choose 8 students from the class of 13.

Now, let's calculate the number of ways to choose 2 female students and 6 male students.

The number of ways to choose 2 female students from 5 is calculated as:

C(5, 2) = 5! / (2!(5-2)!) = 5! / (2!3!) = (5 * 4) / (2 * 1) = 10

Similarly, the number of ways to choose 6 male students from 8 is calculated as:

C(8, 6) = 8! / (6!(8-6)!) = 8! / (6!2!) = (8 * 7) / (2 * 1) = 28

Now, we can calculate the probability:

Probability = (Number of desired outcomes) / (Total number of outcomes)

Number of desired outcomes = Number of ways to choose 2 female students and 6 male students = C(5, 2) * C(8, 6) = 10 * 28 = 280

Total number of outcomes = Number of ways to choose 8 students from 13 = C(13, 8) = 1287

Probability = 280 / 1287 = 0.2174 (rounded to 4 decimal places)

Therefore, the probability that 2 female students and 6 male students will be selected for the oral exam is approximately 0.2174.